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Varvara68 [4.7K]
1 year ago
12

Which is a zero of the quadratic function f(x) = 16x2 32x − 9?

Mathematics
2 answers:
tankabanditka [31]1 year ago
8 0
A zero of a function is the value of x for which f(x) = 0
f(x) = 16x^2 + 32x - 9 = 0
16x^2 - 4x + 36x - 9 = 0
(16x^2 - 4x) + (36x - 9) = 0
4x(4x - 1) + 9(4x - 1) = 0
(4x + 9)(4x - 1) = 0
4x + 9 = 0 and 4x - 1 = 0
4x = -9 and 4x = 1
x = -9/4 and x = 1/4
Lunna [17]1 year ago
7 0

Solution:

The given Quadratic function whose zero we have to find is:

  16 x^2 - 32 x -9=0\\\\16x^2-36 x + 4 x -9=0\\\\4 x\times (4 x -9)+1 \times(4 x -9)=0\\\\(4 x +1)(4 x-9)=0\\\\4 x +1=0 \text{or} ,4 x -9=0\\\\ x=\frac{-1}{4} \text{or} , x=\frac{9}{4}

So,

     x=\frac{-1}{4} \text{or} , x=\frac{9}{4}, are zero of Quadratic function.

If the equation was like this

\rightarrow 16 x^2 + 32 x -9=0\\\\16 x^2+36 x -4 x -9=0\\\\4 x\times (4 x +9)-1 \times(4 x +9)=0\\\\(4 x -1)(4 x+9)=0\\\\4 x -1=0 \text{or} ,4 x +9=0\\\\ x=\frac{1}{4} \text{or} , x=\frac{-9}{4}

So,

     x=\frac{1}{4} \text{or} , x=\frac{-9}{4}, are zero of Quadratic function.

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The blue M&M was introduced in 1995. Before then, the color mix in a bag of plain M&Ms was (30% Brown, 20% Yellow, 20% R
IRISSAK [1]

Answer:

The probability that the yellow M&M came from the 1994 bag is 0.07407 or 7.407%

Step-by-step explanation:

Given

Before 1995

(Br) Brown = 30%

(Y) Yellow = 20%  =0.2

(R) Red = 20%

(G) Green =10%  =0.1

(O) Orange = 10%

(T) Tan = 10%

 

After 1995

(Br) Brown = 13%

(Y) Yellow = 14%  =0.14

(R) Red = 13%

(G) Green = 20% = 0.2

(O) Orange = 16%

(Bl) Blue = 24%

Since there are two bags, let A be the bag from 1994, and B be the bag from 1996

Then let AY imply we drew a yellow M&M from the 1994 bag

AG implies we drew a green M&M from the 1994 bag

BY implies imply we drew a yellow M&M from the 1996 bag

BG implies we drew a green M&M from the 1996 bag

P(AY) =0.2

P (BY) = 0.14

P(AG) =0.1

P(BG) =0.2

Since the draws from the 1994 and 1996 bag are independent,

therefore

P(AY n BG) = 0.2 * 0.2 = 0.04  -------(1)\\P(AG n BY) =0.1 * 0.14 =0.014   --------(2)\\

The draws can happen in either of the 2 ways in (1) and (2) above

therefore total probability E is given as

E =P( AY n BG) u P(AG n BY)\\=0.04 + 0.014 =0.O54

For the yellow one to be from 1994, it implies that the event to be chosen is

P(AYnBG) = 0.2*0.2

Since the total probability is given as E=0.054

then P((AYnBG) /E) =\frac{0.04}{0.054} = 0.07407

Concluding statement: This is the condition for the Yellow one to come from 1994 and green from 1996 provided that they obey the condition from E

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Answer:

1. V = D/t => D=Vt. as V = 16mph

D = 16t

2. Prya has been riding for t+½ hours

3. Prya has gone 12(t+½) miles

4. When Prya and Han meet

16t = 12t +6

4t = 6

t = 3/2 hours

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denis-greek [22]
Since our sum is 62, and we have consecutive numbers being added, our range is most likely going to be from 10-20.
To further solve this problem easily, try finding 4 consecutive numbers with the ones place adding up to 12, 22, 32, 42, or any number that sums up to the 2 in the ones place.
Let's try 14, 15, 16, 17, since 4, 5, 6, and 7 sum up to a number with a 2 in the ones place.
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We have found the age of the friends, but we're looking for the oldest.
Our largest number is 17.
Your answer is 17 years old.
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